Given $(A \vee B) \vee C = A \vee B \vee C = A \vee (B \vee C)$ is true because of the associative law of linear algebra. I have a question about this rule in the case of a negative.

I have: $\neg(A \vee (B \wedge C)) \vee (B \wedge C)$

Treating the $(B \wedge C)$ as a single term, am I correct to apply the associative rule in this case resulting in $\neg A \vee ((B \wedge C) \vee (B \wedge C))$?

Checking my intial expression with the resulting one with an online simplifier tells me they are both equal, but I am not sure why that is.

  • $\begingroup$ Later includes tautology no ? $\endgroup$ – user645636 Jan 21 '20 at 18:58

You cannot apply the associative property here. The negator encloses the whole of $(A \vee (B\land C))$ so you do not actually have an expression of the form $X\vee Y \vee Z$. Coincidentally, you did get the correct result but it would be derived as follows:

\begin{align*} &\neg(A \vee (B \land C)) \vee (B \land C)\\ =&(\neg A \land \neg(B\land C)) \vee (B\land C) \\ =& (\neg A \vee (B\land C)) \land (\neg (B\land C) \vee (B\land C))\\ =& \neg A \vee (B\land C). \end{align*}

using the distributive properties for "$\neg$", as well as for "$\vee$ and $\land$".


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